Math 25: Solutions To Homework # 3 Worksheet - Dartmouth College Page 2
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1
2
3
4(4.1 # 12) Construct a table for addition modulo 6.
+ 0 1 2 3 4 5
0 0 1 2 3 4 5
1 1 2 3 4 5 0
2 2 3 4 5 0 1
3 3 4 5 0 1 2
4 4 5 0 1 2 3
5 5 0 1 2 3 4
(4.1 # 14) Construct a table for multiplication modulo 6.
0 1 2 3 4 5
0 0 0 0 0 0 0
1 0 1 2 3 4 5
2 0 2 4 0 2 4
3 0 3 0 3 0 3
4 0 4 2 0 4 2
5 0 5 4 3 2 1
(4.1 # 20) Show that if n is an odd positive integer or if n is a positive integer divisible by
4, then
3
3
3
1
+ 2
+
+ (n
1)
0 (mod n).
Is this statement true if n is even but not divisible by 4?
By a problem from the first HW,
2
2
2
n(n
1)
n
(n
1)
3
3
3
1
+ 2
+
+ (n
1)
=
=
.
2
4
If 4 n, then n = 4k for some integer k, so
2
2
n
(n
1)
2
= kn(n
1)
0 (mod n).
4
If n is odd then n
1 is even, so n
1 = 2m for some integer m. Then
2
2
n
(n
1)
2
2
= n
m
0 (mod n).
4
If n is even but not divisible by 4, then n = 2 for some odd integer , and
2
2
n
(n
1)
2
2
2
2
2
2
2
=
(n
1)
=
n
2
n +
(mod n),
4
2
2
and since
is odd and n is even, n
, so
0 (mod n).
(4.1 # 22) Show by induction that if n is a positive integer, then 4
1 + 3n (mod 9).
For the base case, 4
1+3 (mod 9). For the induction hypothesis, assume that 4
1+3n
(mod 9) for some positive integer n. Then
+1
4
= 4 4
4(1 + 3n)
4 + 12n
4 + 3n
1 + 3(n + 1) (mod 9).
Therefore 4
1 + 3n (mod 9) for all positive integers n.
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